As a certain object falls, its position s (in meters) above ground after t secon

Question

As a certain object falls, its position s (in meters) above ground after t seconds is given by

s(t) = 30 5t2.

(a) What is the average velocity of the object on the interval from t = 1 to the time 0.5 seconds later?
m/s

(b) What is the average velocity of the object on the interval from t = 1 to the time 0.1 seconds later?
m/s

(c) Write a polynomial in terms of h for the average velocity from t = 1 to the time h seconds later

(h 0).

(Simplify your answer completely.)

(d) What does this average tend toward for h closer and closer to 0 (smaller and smaller time interval)?
m/s

(e) What would you guess to be the exact velocity of the object after 1 second?
m/s

Explanation / Answer

We have given certain object's position s(t) = 30 5t2 in meters after t seconds

a) we have t=1 to 0.5 seconds

the average velocity of the object on the interval from t = 1 to the time 0.5 is [s(1.5)-s(1)]/(1.5-1)

[s(1.5)-s(1)]/(1.5-1)=[(30 5(1.5)2)-(30 5(1)2)]/0.5

=[18.75-25]/0.5

=-12.5 m/s

the average velocity of the object on the interval from t = 1 to the time 0.5 is -12.5 m/s

b) we have t=1 to 0.1 seconds

the average velocity of the object on the interval from t = 1 to the time 0.1 is [s(1.1)-s(1)]/(1.1-1)

[s(1.1)-s(1)]/(1.1-1)=[(30 5(1.1)2)-(30 5(1)2)]/0.1

=[23.95-25]/0.1

=-10.5

the average velocity of the object on the interval from t = 1 to the time 0.1 is -10.5 m/s

c) we have t=1 to the time h seconds

the average velocity from t = 1 to the time h seconds is [s(1+h)-s(1)]/(1+h-1)

[s(1+h)-s(1)]/(1+h-1)=[30 5(1+h)2-(30 5(1)2)]/(1+h-1)

=[30-5-5h^2-10h-25]/h

=h(-5h-10)/h

=-5h-10 since h get cancelled

the average velocity from t = 1 to the time h seconds is -5h-10

d) plug h=0 into part (c)

the average tend toward for h closer and closer to 0 (smaller and smaller time interval) is -10 m/s

e) we have s(t)=30-5t^2

v(t)=s'(t)=-10t

after 1 second velocity is v(1)=-10*1=-10 m/s

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